hdu 1028 Ignatius and the Princess III

Ignatius and the Princess III

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 32   Accepted Submission(s) : 28

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Problem Description

"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

"The second problem is, given an positive integer N, we define an equation like this:
  N=a[1]+a[2]+a[3]+...+a[m];
  a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
  4 = 4;
  4 = 3 + 1;
  4 = 2 + 2;
  4 = 2 + 1 + 1;
  4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

Input

The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

Output

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

Sample Input

41020

Sample Output

542
627
和凑数字很像，问一个数分解不考虑顺序，有多少种分法可以用母函数做（母函数要慢一些），也可以用多重背包的思想去做，只不过这里的背包不是取最大了，而是加上之前的状态的种类数，注意要让dp[0]=1;
母函数
#include<iostream>
#include<cmath>#include<cstring>#include<algorithm>#include<queue>using namespace std;int main(){    int  a1[1111];    int  a2[1111];    int n;    while(cin>>n)    {       int i,j,k;       a1[0]=1;       a2[0]=0;       for(i=1;i<=n;i++)       {           a1[i]=1;           a2[i]=0;       }       for(i=2;i<=n;i++)       {           for(j=0;j<=n;j++)                         for(k=0;k+j<=n;k+=i)              {                  a2[k+j]+=a1[j];                  }           for(int s=0;s<=n;s++)           {               a1[s]=a2[s];               a2[s]=0;               }                   }       cout<<a1[n]<<endl;    }    return 0;}
背包
#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#include<queue>using namespace std;int main(){      long long int dp[35555];      int n;      while(cin>>n){       memset(dp,0,sizeof(dp));      dp[0]=1;      int i,j;         for(i=1;i<=n;i++)      {          for(j=i;j<=n;j++)          dp[j]+=dp[j-i];      }    cout<<dp[n]<<endl;}        return 0;}